Solución

Las fórmulas para wu\frac{\partial w}{\partial u} y wv\frac{\partial w}{\partial v} son

wu=wxxu+wyyu+wzzu\frac{\partial w}{\partial u} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial u} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial u} wv=wxxv+wyyv+wzzv\frac{\partial w}{\partial v} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial v} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial v}

Por lo tanto, hay nueve derivadas parciales diferentes que deben calcularse y sustituirse. Necesitamos calcular cada una de ellas:

wx=6x2y\frac{\partial w}{\partial x} = 6x-2ywy=2x\frac{\partial w}{\partial y} = -2xwz=8z\frac{\partial w}{\partial z} = 8z
xu=eusenv\frac{\partial x}{\partial u} = e^usenvyu=eucosv\frac{\partial y}{\partial u} = e^ucosvzu=eu\frac{\partial z}{\partial u} = e^u
xv=eucosv\frac{\partial x}{\partial v} = e^ucosvyv=eusenv\frac{\partial y}{\partial v} = -e^usenvzv=0\frac{\partial z}{\partial v} = 0

Ahora, sustituimos cada una de ellas en la primera fórmula para calcular wu\frac{\partial w}{\partial u}:

wu=wxxu+wyyu+wzzu=(6x2y)eusenv2xeucosv+8zeu\begin{aligned} \frac{\partial w}{\partial u} &= \frac{\partial w}{\partial x}\cdot\frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\cdot\frac{\partial y}{\partial u} + \frac{\partial w}{\partial z}\cdot\frac{\partial z}{\partial u}\\ &= (6x − 2y)e^usen v − 2xe^ucos v + 8ze^u \end{aligned}

entonces sustituye x(u,v)=eusenvx (u, v) = e^usen v, y(u,v)=eucosvy (u, v) = e^ucos v, y z(u,v)=euz (u, v) = e^u en esta ecuación:

wu=(6x2y)eusenv2xeucosv+8zeu=(6eusenv2eucosv)eusenv2(eusenv)eucosv+8e2u=6e2usen2v4e2usenvcosv+8e2u=2e2u(3sen2v2sinvcosv+4)\begin{aligned} \frac{\partial w}{\partial u} &= (6x − 2y)e^usen v − 2xe^ucos v + 8ze^u\\ &= (6e^usen v − 2e^ucos v)e^usen v − 2(e^usen v)e^ucos v + 8e^{2u}\\ &= 6e^{2u}sen^2v − 4e^{2u}sen v cos v + 8e^{2u}\\ &= 2e^{2u}\big(3 sen^2v − 2 sin v cos v + 4\big) \end{aligned}

A continuación, calculamos wv\frac{\partial w}{\partial v}:

wv=wxxv+wyyv+wzzv=(6x2y)eucosv2x(eusenv)+8z(0)\begin{aligned} \frac{\partial w}{\partial v} &= \frac{\partial w}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial v} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial v}\\ &= (6x − 2y)e^ucos v − 2x(−e^usen v) + 8z(0) \end{aligned}

entonces sustituimos x(u,v)=eusenvx (u, v) = e^usen v, y(u,v)=eucosvy (u, v) = e^ucos v, y z(u,v)=euz (u, v) = e^u en esta ecuación:

wv=(6x2y)eucosv2x(eusenv)=(6eusenv2eucosv)eucosv+2(eusenv)(eusenv)=2e2usen2v+6e2usenvcosv2e2ucos2v=2e2u(sen2v+senvcosvcos2v)\begin{aligned} \frac{\partial w}{\partial v} &= (6x − 2y)e^ucos v − 2x(−e^usen v)\\ &= (6e^usen v − 2e^ucos v)e^ucos v + 2(e^usen v)(e^usen v)\\ &= 2e^{2u}sen^2v + 6e^{2u}sen v cos v − 2e^{2u}cos^2v\\ &= 2e^{2u}\big(sen^2v + sen v cos v − cos^2v\big) \end{aligned}